can anyone help with this problem?
Prove
tan^2 + cos^2 + sin^2 = 1/cos^2
$\displaystyle \tan^2 x - \cos^2 x= \frac{1}{\cos ^2 x}- 1 -\cos ^2 x$
$\displaystyle (\sec^2 x-1)-\cos^2 x=\frac{1}{\cos^2 x}-1-\cos^2 x$
$\displaystyle \sec^2 x - 1 - \cos^2 x= \frac{1}{\cos^2 x}-1 -\cos^2 x$
$\displaystyle \frac{1}{\cos^2 x}-1-\cos^2 x = \frac{1}{\cos^2 x}-1 -\cos^2 x$
Hello,
you have to use these properties:
$\displaystyle \cos^2(x)+\sin^2(x)=1$
$\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$
$\displaystyle \tan^2(x)+\cos^2(x)+\sin^2(x)=\tan^2(x)+1-\sin^2(x)+\sin^2(x)=$$\displaystyle
\tan^2(x)+1=\frac{\sin^2(x)}{\cos^2(x)}+1=\frac{\s in^2(x)+\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}$
EB
PS: If you have a new problem please start a new thread. Otherwise you risk that nobody will notice that you asked for some help.
I'll reinforce that comment. If you remove your original question then no one can see what you originally posted and thus be confused about what someone's response to it is. If someone else is looking for help on a similar topic, that makes it harder for them to learn it themselves. So please, when you have a new question either post a completely new thread, or just respond to the post you already started.
-Dan